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I am attempting to compute the minimal polynomial of $1 + 2^{1/3} + 4^{1/3}$ over $\mathbb Q$. So far, my reasoning is as follows:

The Galois conjugates of $2^{1/3}$ are $2^{1/3} e^{2\pi i/3}$ and $2^{1/3} e^{4\pi i /3}$. We have $4^{1/3} = 2^{2/3}$, so the image of $4^{1/3}$ under an automorphism $\sigma$ fixing $\mathbb Q$ is determined by the image of $2^{1/3}$: it must equal the square of $\sigma(2^{1/3})$. Therefore, the Galois conjugates of $1 + 2^{1/3} + 4^{1/3}$ are $1 + 2^{1/3} e^{2\pi i/3} + 4^{1/3} e^{4\pi i/3}$ and $1 + 2^{1/3} e^{4\pi i/3} + 4^{1/3} e^{2\pi i/3}$. Therefore, the minimal polynomial is $(x-a)(x-b)(x-c)$, where

$$\begin{align*} a&=1 + 2^{1/3} + 4^{1/3},\\ b&=1 + 2^{1/3} e^{2\pi i/3} + 4^{1/3} e^{4\pi i/3},\text{ and}\\ c&=1 + 2^{1/3} e^{4\pi i/3} + 4^{1/3} e^{2\pi i/3}. \end{align*}$$

However, this polynomial does not seem to have coefficients in $\mathbb Q$! What am I doing wrong?

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Why do you think it doesn't have coefficients in Q? Have you tried multiplying it out? –  NKS Jan 2 '12 at 22:36
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2 Answers 2

up vote 7 down vote accepted

By expanding and using the relation $1+e^{2\pi i/3}+e^{4\pi i/3}=0$ heavily I got that $$ (x-a)(x-b)(x-c)=x^3-3x^2-3x-1. $$ Looks like rational coefficients to me.

Another way of seeing this is to compute $$ (a-1)^3=(2^{1/3}+4^{1/3})^3=2+3\cdot 2^{4/3}+3\cdot 2^{5/3}+4=6+6(2^{1/3}+4^{1/3})=6+6(a-1)=6a. $$ Hence $$ 0=a^3-3a^2+3a-1-6a=a^3-3a^2-3a-1. $$

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We can conclude that your polynomial has rational coefficients even without doing any expanding. The splitting field for the polynomial, which is $\mathbb{Q}(2^{1/3},\omega)$ (here $\omega = e^{2\pi i/3}$) has Galois group $S_3$ generated by the automorphisms $\sigma$ fixing roots of unity and sending $2^{1/3}$ to $2^{1/3}\omega$, and $\tau$ fixing $2^{1/3}$ and sending $\omega$ to $\omega^2$. When we expand $(x-a)(x-b)(x-c)$, the coefficients will be given (up to sign) by the elementary symmetric polynomials in $a,b,c$. Since each element of $S_3$ permutes the set $\{a,b,c\}$, the symmetric polynomials will all be fixed by $S_3 \cong \operatorname{Gal}(\mathbb{Q}(2^{1/3},\omega):\mathbb{Q})$, so will lie in $\mathbb{Q}$.

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